Festo Hydraulic Basic Level Manual

Learning System for Automation

Hydraulics Workbook Basic Level

094468 (04/01)

Authorised applications and liability The Learning System for Automation and Communication has been developed and prepared exclusively for training in the field of automation and communication. The training organization and / or trainee shall ensure that the safety precautions described in the accompanying Technical documentation are fully observed. Festo Didactic hereby excludes any liability for injury to trainees, to the training organization and / or to third parties occurring as a result of the use or application of the station outside of a pure training situation, unless caused by premeditation or gross negligence on the part of Festo Didactic. Order no.: Description: Designation: Edition: Layout: Graphics: Author:

094468 TEACHW. HYDRAUL. D.S501-C-SIBU-GB 04/01 30.04.2001, OCKER Ingenieurbüro OCKER Ingenieurbüro D. Waller, H. Werner

© Copyright by Festo Didactic GmbH & Co., D-73770 Denkendorf 2001 The copying, distribution and utilization of this document as well as the communication of its contents to others without expressed authorization is prohibited. Offenders will be held liable for the payment of damages. All rights reserved, in particular the right to carry out patent, utility model or ornamental design registrations. Parts of this training documentation may be duplicated, solely for training purposes, by persons authorised in this sense.

TP501 · Festo Didactic

3

Preface Festo Didactic’s Learning System for Automation and Communications is designed to meet a number of different training and vocational requirements. The Festo Training Packages are structured accordingly:

Basic Packages provide fundamental knowledge on a wide range of technologies.

Technology Packages deal with important areas of open-loop and closed-loop control technology.

Function Packages explain the basic functions of automation systems.

Application Packages provide basic and further training closely oriented to everyday industrial practice. Technology Packages deal with the technologies of pneumatics, electropneumatics, programmable logic controllers, automation with PCs, hydraulics, electrohydraulics, proportional hydraulics and application technology (handling). Fig. 1: Hydraulics 2000 – i.e. mobile workstation

Mounting frame

U = 230V~

Profile plate

p = 6 MPa

Storage tray

TP501 • Festo Didactic

4

The modular structure of the Learning System permits applications to be assembled which go beyond the scope of the individual packages. It is possible, for example, to use PLCs to control pneumatic, hydraulic and electrical actuators. All training packages have an identical structure:

Hardware Courseware Software Courses

The hardware consists of industrial components and installations, adapted for didactic purposes. The courseware is matched methodologically and didactically to the training hardware. The courseware comprises:

Textbooks (with exercises and examples) Workbooks (with practical exercises, explanatory notes, solutions and data sheets)

OHP transparencies and videos (to bring teaching to life) Teaching and learning media are available in several languages. They have been designed for use in classroom teaching but can also be used for self-study purposes. In the software field, computer-based training programs and programming software for programmable logic controllers are available. Festo Didactic’s range of products for basic and further training is completed by a comprehensive selection of courses matched to the contents of the technology packages.


5

Latest information about the technology package TP501. New in Hydraulic 2000:

Industrial components on the profile plate. Exercises with exercise sheets and solutions, leading questions. Fostering of key qualifications: Technical competence, personal competence and social competence form professional competence.

Training of team skills, willingness to co-operate, willingness to learn, independence and organisational skills. Aim – Professional competence

Content Part A

Course

Exercises

Part B

Fundamentals

Reference to the text book

Part C

Solutions

Function diagrams, circuits, descriptions of solutions and quipment lists

Part D

Appendix

Storage tray, mounting technology and datasheets


6


7

Table of contents Introduction

11

Notes on safety

13

Notes on operation

13

Technical notes

14

Training contents

17

Equipment set for “Hydraulics Basic Level”

19

Component / exercise table for TP 501

24

Section A – Course Exercise 1: Automatic lathe Pump characteristic

A-3

Exercise 2: Package lifting device Pressure relief valve characteristic

A-7

Exercise 3: Drawing press Hydraulic resistances

A-11

Exercise 4: Calender feeding device Single-acting cylinder (basic circuit)

A-15

Exercise 5: Hardening furnace Single-acting cylinder (measurement and calculation)

A-19

Exercise 6: Furnace door control Double-acting cylinder

A-23

Exercise 7: Conveyor tensioning device 4/3-way valve with bypass to pump

A-29

Exercise 8: Cold-store door Accumulator

A-33

Exercise 9: Rotary machining station Flow control valve and counter-holding

A-37

Exercise 10: Painting booth Flow control valve characteristic

A-41


8

Exercise 11: Embossing machine One-way flow control valve and counter-holding

A-45

Exercise 12: Surface grinding machine Differential circuit

A-49

Exercise 13: Drilling machine Pressure regulator

A-55

Exercise 14: Bulkhead door Hydraulic clamping of a cylinder

A-59

Exercise 15: Ferry loading ramp Flow control valve in inlet and outlet lines

A-63

Exercise 16: Skip handling Varying load

A-69

Exercise 17: Bonding press Comparison of pressure regulator and – pressure relief valve

A-73

Exercise 18: Assembly device Pressure sequence circuit, displacement-step diagram

A-77

Exercise 19: Assembly device Calculation of pressure and time

A-81

Exercise 20: Tipping container Electrohydraulics

A-85

Section B - Fundamentals


9

Section C – Solutions Solution 1:

Automatic lathe

C-3

Solution 2:

Package lifting device

C-7

Solution 3:

Drawing press

C-11

Solution 4:

Calender feeding device

C-15

Solution 5:

Hardening furnace

C-19

Solution 6:

Furnace door control

C-23

Solution 7:

Conveyor tensioning device

C-27

Solution 8:

Cold-store door

C-33

Solution 9:

Rotary machining station

C-37

Solution 10:

Painting booth

C-41

Solution 11:

Embossing machine

C-45

Solution 12:

Surface grinding machine

C-49

Solution 13:

Drilling machine

C-59

Solution 14:

Bulkhead door

C-65

Solution 15:

Ferry loading ramp

C-69

Solution 16:

Skip handling

C-73

Solution 17:

Bonding press

C-77

Solution 18:

Assembly device

C-79

Solution 19:

Calculation for an assembly device

C-83

Solution 20:

Tipping container

C-85

Section D – Appendix Storage tray

D-3

Mounting systems

D-4

Sub-base

D-6

Coupling system

D-7

Data sheets


...

10


11

Introduction This workbook forms part of Festo Didactic’s Learning System for Automation and Communications. The Technology Package “Hydraulics”, TP500, is designed to provide an introduction to the fundamentals of hydraulic control technology. This package comprises a basic level and an advanced level. The basic level package TP501 teaches basic knowledge of hydraulic control technology, while the advance level package TP502 builds on this. The basic level hydraulic exercises are designed to be carried out with manual actuation. It is, however, also possible to use electrical actuation. The hydraulic components have been designed to provide the following:

Easy handling Secure mounting Environmentally-friendly coupling system Compact component dimensions Authentic measuring methods

We recommend the following for the practical execution of the exercises:

Hydraulic components: Equipment set TP501 One hydraulic power pack A number of hose lines A profile plate or a suitable laboratory trolley A measuring set with the appropriate sensors


12

This workbook provides knowledge of the physical interrelationships and the most important basic circuits in hydraulics. The exercises deal with the following:

Plotting of characteristics for individual components Comparison of the use of different components Assembly of various basic circuits Use of basic hydraulics equations

The following technical equipment is required for safe operation of the components:

A hydraulic power pack providing an operating pressure of 60 bar and a flow rate of 2 l/min

An electrical power supply of 230V AC for the hydraulic power pack A power supply unit with an output of 24V DC for solenoid-actuated valves

A Festo Didactic profile plate for mounting the components The theoretical background is described in the “Hydraulics Basic Level” textbook TP501. Technical descriptions of the components used are given in the data sheets in Part D of this workbook. Festo Didactic offers the following further training material for hydraulics:

Magnetic symbols Hydraulics slide rule Set of OHP transparencies Transparent models Interactive video Symbol library


13

Notes on safety Observe the following in the interests of your own safety:

Exercise care when switching on the hydraulic power pack. Cylinders may advance unexpectedly!

Do not exceed the maximum permissible operating pressure (see data sheets).

Observe all general safety instructions (DIN 58126 and VDE 100).

Notes on operation Always work in the following sequence when assembling a hydraulic circuit. 1. The hydraulic power pack and electrical power supply must be switched off during the assembly of the circuit. 2. All components must be securely fitted to the profile plate, i.e. securely snap-fitted or bolted down. 3. Check that all return lines are connected and all hose lines are securely fitted. 4. Switch on the electrical power supply first and then the hydraulic power pack. 5. Before dismantling the circuit, ensure that pressure in hydraulic components has been released: Couplings must be connected and disconnected only under zero pressure! 6. Switch off the hydraulic power pack first and then the electrical power supply.


14

Technical notes Observe the following in order to ensure safe operation.

The hydraulic power pack PN 152962 incorporates an adjustable pressure relief valve. In the interests of safety, the pressure is limited to approx. 60 bar (6 MPa).

The maximum permissible pressure for all hydraulic components is 120 bar (12 MPa). The operating pressure should not exceed 60 bar (6 MPa).

In the case of double-acting cylinders, the pressure intensification effect may produce an increased pressure proportional to the area ratio of the cylinder. With an area ratio of 1:1.7 and an operating pressure of 60 bar (6 MPa), this increased pressure may be over 100 bar (10 MPa)! Fig. 2: Pressure intensification

If connections are detached under pressure, the non-return valve in the coupling may cause pressure to become trapped in the valve or other component concerned. The pressure relieving device PN 152971 can be used to release this pressure. Exception: This is not possible in the case of hose lines and non-return valves.

All valves, other components and hose lines are fitted with selfclosing quick-release couplings. This prevents the accidental spillage of hydraulic fluid. In the interests of simplicity, these couplings are not shown in circuit diagrams. Fig. 3: Simplified drawing of self-closing couplings

Throttle valve

Hose

Shut-off valve


15

It is frequently necessary when assembling a control circuit to modify the given circuit diagram. Within the scope of the equipment set in this Training Package, the following alternative solutions are possible:

Plugs can be used to change the function of directional control valves (Figs. 4 and 5).

Directional control valves with different normal positions can be used (Fig. 6).

Solenoid-actuated valves can be used in place of hand lever valves (Fig. 7). 2/2-way valve

3/2-way valve

Fig. 4: Circuit diagram

4/2-way valve

4/2-way valve

Fig. 5: Practical assembly

Circuit diagram

Practical assembly

Fig. 6: Directional control valves with various normal positions

Fig. 7: Solenoid-actuated directional control valve


16

Flow rate sensor The flow rate sensor consists of:

A hydraulic motor, which converts the flow rate q into a rotary speed n.

A tachogenerator, which produces a voltage V proportional to the rotary speed n.

A universal display unit, which converts the flow rate q into l/min. The universal display should be set to sensor no. 3 for this purpose. Fig. 8: Block circuit diagram

q

Hydraulic motor

n

Tachogenerator

V

Universal display

q

Fig. 9: Circuit diagrams, hydraulic and electrical

Fig. 10: Connecting up the universal display Battery operation

External power supply


17

Training contents Characteristics of valves and other components. Uses of individual valves and other components. Comparison of uses and functions of different valves and other components.

Measurement of variables such as pressure, flow rate and time. Control of pressure and speed. Calculations of area ratios, forces, power and speed. Basic physical principles of hydraulics. Use of basic hydraulics equations. Understanding and drafting of circuit diagrams. Drafting of displacement/step diagrams. Use of symbols in accordance with DIN/ISO 1219. Assembly and commissioning of control circuits, including faultfinding.

Assessment of energy consumption. Basic hydraulic circuits such as a pressure sequence circuit, a bypass circuit to the pump, a differential circuit, circuits with flow control valves in the inlet, outlet and bypass, circuits with counter-holding and bypass circuits with a non-return valve.


18

List of training aims

Exercise

Training aims

1

Drawing a pump characteristic.

2

Drawing a characteristic for a pressure relief valve.

3

Measuring flow resistances.

4

Application of a non-return valve. Use of a 2/2-way valve to control a single-acting cylinder.

5

Application of a 3/2-way valve. Determination of times

6

Application of a 4/2-way valve. Determination of times

7

Application of a 4/3-way valve. Use of a pilot-operated non-return valve.

8

Use of a hydraulic accumulator as a power source. Use of accumulator to power advance and return strokes of cylinder after pump is switched off.

9

Application of a 2-way flow control valve. Assembly of a counter-pressure circuit.

10

Plotting of characteristic for a 2-way flow control valve. Comparison between this valve and a throttle valve.

11

Application of a one-way flow control valve. Difference between flow control valve and throttle valve on the basis of a concrete application.

12

Design and mode of operation of a differential circuit. Influence of piston areas on pressures

13

Design of a control circuit with reduced output pressure. Explanation of mode of operation of a 3-way pressure regulator.

14

Hydraulic clamping with a double-acting cylinder. Comparison of circuits with and without counter-holding.

15

Speed control circuit with tractive load. Comparison of circuits with flow control valves in the inlet line and outlet line respectively.

16

Circuit for a double-acting cylinder with a varying load.

17

Specification of pressure for a double-acting cylinder. Choice of either a pressure relief valve or a pressure regulator

18

Pressure sequence circuit. Drawing of a displacement/step diagram

19

Calculation of forces associated with a double-acting cylinder Calculation of advance-stroke time of a cylinder piston.

20

Electrohydraulic control circuit.


19

Equipment set for “Hydraulics Basic Level” Description

Order No.

Qty.

Pressure gauge

152841

3

Throttle valve

152842

1

One-way flow control valve

152843

1

Shut-off valve

152844

1

Non-return valve, opening pressure 1 bar

152845

1

Non-return valve, opening pressure 5 bar

152846

1

Branch tee

152847

7

Pressure relief valve

152848

1

Pressure relief valve, piloted

152849

1

Pressure regulator

152850

1

Flow control valve

152851

1

Non-return valve, hydraulically piloted

152852

1

Double-acting cylinder

152857

1

Hydraulic motor

152858

1

Diaphragm accumulator

152859

1

Loading weight, 9 kg

152972

1

4/2-way hand lever valve

152974

1

4/3-way hand lever valve, recirculation mid-position

152977

1


TP501, PN 080246

20

Additional equipment

Accessories

Description

Order No.

Qty.

Stop-watch

151504

1

4/3-way hand lever valve, closed in mid-position

152975

1

4/3-way hand lever valve, relieving mid-position

152976

1

Relay, 3-fold

162241

1

Signal input unit, electrical

162242

1

Flow-rate sensor

183736

1

4/2-way solenoid valve

167082

1

4/3-way solenoid valve, closed in mid-position

167083

1

4/3-way solenoid valve, relieving mid-position

167084

1

4/3-way solenoid valve, recirculating mid-position

167085

1

Universal display

183737

1

Pressure sensor

184133

1

Order No.

Qty.

Profile plate, large

159411

1

Schlauchleitung, 600 mm

152960

12

Hydraulik-Aggregat

152962

1

Hose line, 1000 mm

152970

4

Pressure relieving device

152971

1

Protective cover (for weight, 9kg)

152973

1

Power supply unit, 24 V, 4.5 A

162417

1

Cable set with safety plugs

167091

1

Description


21

Description Pressure gauge

Throttle valve


Shut-off valve Non-return valve Branch tee


Pilot-operated pressure relief valve

Pressure regulator

Flow control valve

Piloted non-return valve

Double-acting cylinder


Symbol

Symbols for equipment set TP501

22


Description

Symbol

Hydraulic motor

Diaphragm accumulator, detailed

Diaphragm accumulator, simplified

Weight 4/2-way hand lever valve


4/3-way hand lever valve, closed in mid-position

4/3-way hand lever valve, relieving mid-position

4/3-Wege-Handhebelventil mit Umlaufstellung


23

Description 4/3-way solenoid valve, closed in mid-position

4/3-way solenoid valve, relieving mid position

4/3-way solenoid valve, recirculating mid-position

Hose line

Hydraulic-power pack, detailed

Hydraulic power pack, simplified

Pressure sensor

Flow rate sensor

Hydraulic motor with tachogenerator


Symbol


24

Component / exercise table for TP 501 Exercises Description

1

2

3

4

5

6

7

8

9

10 11 12 13 14 15 16 17 18 19 20

Pressure gauge

1

1

3

1

3

1

1

2

5

3

Throttle valve

1 1 1

1

1

1

1

Non-return valve, 1 bar 1

1

1 1

1

Non-return valve, 5 bar

3

4

3

3

3

3

2

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

One-way flow control valve Shut-off valve

5

1

1

1

1

1

1

Branch tee

2

3

4

3

2

3

3

6

2

4

4

5

4

4

4

5

7

2

Pressure relief valve *)

1

2

1

1

1

1

1

2

2

2

1

1

2

2

3

2

3

1

Pressure relief valve, piloted

(1) (1)

(1) (1) (1)

(1) (1) (1) (1) (1)

Pressure regulator

1

Flow control valve

1

1

Piloted non-return valve

1

1

1

1

1

1

Cylinder, double-acting

1

1

1

1

1 1

1

1

1

1

1

1

1

1

Hydraulic motor

1

1

1


1

Weight

1

4/2-way hand lever valve

1

4/3-way hand lever valve recirculating mid-position

1

1 1

1 1

1 1

1

1

1

1

1 1

1

1

1

Hydraulic power pack

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

Hose line, 600 mm

3

5

5

6

5

4

7

5

9

4

12

5

12 10 11

8

5

12

4

2

2

2

2

2

2

2

2

2

2

2

2

2

4

2

1

1

1

1

1

2

2

Hose line, 1000 mm Stop-watch Pressure sensor

(2)

1

1

1

2

2

1

1

Flow-rate sensor

1

1

1

1

1

Universal display

1

1

1

1

1

Power supply unit

1

1

1

1

1

1

*) If a sufficient number of directly-controlled pressure relief valves is not available, the pilot-operated pressure relief valve can also be used.


25

Electrical equipment for exercise 20 Description

Order No.

Qty.

4/3-way solenoid valve, relieving mid-position

167084

1

Signal input unit, electrical

162242

1

Relay, 3-fold

162241

1

Cable set

167091

1

The exercises appear in Section A of the workbook, with solutions to these in Section C. The methodological structure is the same for all exercises. The exercises in Section A are structured as follows:

Subject Title Training aim(s) Problem definition Exercise Positional sketch

A worksheet then follows for use in carrying out the exercise. The solutions in Section C contain the following:

Hydraulic circuit diagram Practical assembly Component list Solution description Evaluation Conclusions


Methodological structure of exercises

26


A-1

Section A – Course Exercise 1: Automatic lathe Pump characteristic

A-3

Exercise 2: Package lifting device Pressure relief valve characteristic

A-7

Exercise 3: Drawing press Hydraulic resistances

A-11

Exercise 4: Calender feeding device Single-acting cylinder (basic circuit)

A-15

Exercise 5: Hardening furnace Single-acting cylinder (measurement and calculation)

A-19

Exercise 6: Furnace door control Double-acting cylinder

A-23

Exercise 7: Conveyor tensioning device 4/3-way valve with bypass to pump

A-29

Exercise 8: Cold-store door Accumulator

A-33

Exercise 9: Rotary machining station Flow control valve and counter-holding

A-37

Exercise 10: Painting booth Flow control valve characteristic

A-41

Exercise 11: Embossing machine One-way flow control valve and counter-holding

A-45

Exercise 12: Surface grinding machine Differential circuit

A-49

Exercise 13: Drilling machine Pressure regulator

A-55

Exercise 14: Bulkhead door Hydraulic clamping of a cylinder

A-59


A-2

Exercise 15: Ferry loading ramp Flow control valve in inlet and outlet lines

A-63

Exercise 16: Skip handling Varying load

A-69

Exercise 17: Bonding press Comparison of pressure regulator and – pressure relief valve

A-73

Exercise 18: Assembly device Pressure sequence circuit, displacement-step diagram

A-77

Exercise 19: Assembly device Calculation of pressure and time

A-81

Exercise 20: Tipping container Electrohydraulics

A-85


A-3 Exercise 1

Hydraulics

Subject

Automatic lathe

Title

To teach the student how to draw the characteristic curve for a pump

Training aim

Drawing the hydraulic circuit diagram Practical assembly of the circuit Determining the various measured values and entering them into the

Problem definition

table

Drawing the characteristic curve for the pump Drawing conclusions


A-4 Exercise 2

Exercise

The main spindle on an automatic lathe is driven by a hydraulic motor, while a hydraulic cylinder is used to execute a feed movement of the workpiece slide. It has been established that the specified speed is no longer reached during the processing cycle. The pump characteristic curve is therefore to be evaluated.

Positional sketch


A-5 Exercise 1

EXERCISE SHEET

System pressure p

15

20

25

30

35

40

45

Flow rate q

50

bar

Evaluation

l/min

Pump characteristic

How does the flow rate change as the pressure increases?


Conclusion

A-6 Exercise 2


A-7 Exercise 2

Hydraulics

Subject


Title

To teach the student how to draw the characteristic for a

Training aim

pressure relief valve

Drawing the hydraulic circuit diagram Practical assembly of the circuit Setting a maximum pressure of 50bar Establishing the opening pressure of the pressure relief valve Determining the various measured values and entering them into the table

Drawing the pressure/flow rate characteristic Drawing conclusions


Problem definition

A-8 Exercise 2

Exercise

Owing to a change in the production process, a package lifting device is now required to lift heavier packages than those for which it was originally designed. It has been observed that the stroke speed is now lower. Using the pressure/flow rate characteristic for the pressure relief valve, determine the pressure at which flow diversion of the pump output begins.

Positional sketch

1A


A-9 Exercise 2

EXERCISE SHEET

Working pressure p Flow rate q

35

40

42.5

45

47.5

50

bar

Evaluation

l/min

Characteristic for pressure relief valve

How great is the difference between the opening pressure and maximum pressure?


Conclusion

A-10 Exercise 2


A-11 Exercise 3

Hydraulics

Subject

Drawing press

Title

To teach the student how to measure flow resistances

Training aim

Problem definition

Drawing the hydraulic circuit diagram Practical assembly of the circuit Setting a constant flow rate Measuring the flow resistances Drawing conclusions


A-12 Exercise 3

Exercise

A drawing press is used to shape metal workpieces. Following modification of the hydraulic system, the workpieces are no longer dimensionally accurate. The reason for this may be that the required press pressure is not being reached. Use a special test set-up to measure the hydraulic resistance of the components used.

Positional sketch


A-13 Exercise 3

EXERCISE SHEET

Evaluation

p0Z3 = Pressure upstream of component p0Z4 = Pressure downstream of component Component

Flow rate q l/min

Pressure relief valve, fully open Throttle valve, fully open 4/2-way valve, P –> A 4/3-way valve, P –> A

Pressure p0Z3 bar

Pressure p0Z4 bae

Pressure difference ∆p bar

2 1 2 1 2 1 2 1

How does the pressure difference change when the flow rate is doubled?


Values table

Conclusion

A-14 Exercise 3


A-15 Exercise 4

Hydraulics

Subject


Title

To familiarise the student with the applications of a non-return valve To show the activation of a single-acting cylinder using a 2/2-way

Training aim

valve

Drawing the hydraulic circuit diagram Practical assembly of the circuit Evaluation of this circuit


Problem definition

A-16 Exercise 4

Exercise

Rolls of paper are lifted into a calender by a lifting device. The lifting device is driven by a plunger cylinder (single-acting cylinder). When the hydraulic power pack is switched on, the pump output flows directly to the cylinder. A 2/2-way valve, which is closed in its normal position, is fitted in a branch line leading to the tank. A non-return valve is used to ensure that the pump is protected against the oil back-pressure. A pressure relief valve is fitted upstream of the non-return valve to safeguard the pump against excessive pressures.

Positional sketch


A-17 Exercise 4

EXERCISE SHEET

Circuit diagram, hydraulic


A-18 Exercise 4

Components list

Conclusion

Item no.

Qty.

Description

What is the disadvantage of this circuit?


A-19 Exercise 5

Hydraulics

Subject

Hardening furnace

Title

To familiarise the student with the applications of a 3/2-way valve To show how to determine times, pressures and forces during the

Training aim

advance and return strokes of a single-acting cylinder

Drawing the hydraulic circuit diagram Determining the necessary components Practical assembly of the circuit Measuring the travel pressure and travel time for the advance and return strokes

Calculating the required advance-stroke pressure Calculating the advance-stroke speed and time


Problem definition

A-20 Exercise 5

Exercise

The cover of a hardening furnace is to be raised by a single-acting cylinder. The cylinder is activated by a 3/2-way valve. A 9 kg weight is attached to the cylinder to represent the load. Measure and calculate the following values:

Travel pressure, load pressure, resistances and back pressure Advance-stroke time and speed Positional sketch


A-21 Exercise 5

EXERCISE SHEET

Direction

Travel pressure

Travel time

Evaluation

Advance stroke Return stroke

Characteristic data required for calculation: Applied load:

FG = 90 N

Piston area:

APN = 2 cm

2

Stroke length:

s = 200 mm

Pump output:

q = 2 l/min

Load pressure:

pL =

FG A PN

pL = Hydraulic resistance = Travel pressure - load pressure p res =

How great is the back pressure in relation to the hydraulic resistance?


Conclusion

A-22 Exercise 5

Advance-stroke speed: v adv =

q A PN

v adv =

Advance-stroke time:

t adv =

s v adv

t adv =

Conclusion

Do the calculated and measured advance-stroke times agree?


A-23 Exercise 6

Hydraulics

Subject


Title

To familiarise the student with the applications of a 4/2-way valve To show how to determine times, pressures and forces during the

Training aim

advance and return strokes of a double-acting cylinder

Drawing the hydraulic circuit diagram Determining the necessary components Practical assembly of circuit Measuring the travel and back pressures and transfer time for the advance and return strokes

Calculation of advance and return-stroke speeds Comparison of calculated and measured values


Problem definition

A-24 Exercise 6

Exercise

A furnace door is opened and closed by a double-acting cylinder. The cylinder is activated by a 4/2-way valve with spring return. This ensures that the door opens only as long as the valve is actuated. When the valve actuating lever is released, the door closes again.

Positional sketch


A-25 Exercise 6

EXERCISE SHEET



A-26 Exercise 6

Evaluation

Advance stroke

Travel pressure p1S1

Back pressure p1S2

Travel time tadv

Return stroke

Back pressure p1S1


Travel time tret

Characteristic data required for calculation: 2

Piston area:

APN = 2.0 cm

Piston annular area:

APR = 1.2 cm

2

Stroke length:

s = 200 mm

Pump output:

q = 2 l/min

Area ratio:

α=

A PN A PR

α=

Advance-stroke speed.:

v adv =

q A PN

Vadv =


t adv =

s v adv

t adv =

Return-stroke speed:

v ret =

q A PR

Vret =


A-27 Exercise 6

Return-stroke time:

t ret =

s v ret

t ret =

Ratio of travel speeds: Vadv = Vret

Ratio of travel times: t adv = t ret

Compare the advance- and return-stroke speeds and times with the area ratio. What is the relationship between these?


Conclusion

A-28 Exercise 6


A-29 Exercise 7

Hydraulics

Subject


Title

To familiarise the student with the applications of a 4/2-way valve To show how to use a piloted non-return valve

Training aim

Problem definition

Drawing the hydraulic circuit diagram Determining the necessary components Practical assembly of the circuit Measuring travel and back pressure and the system pressure in all valve positions

Calculating the power balance for circuits with various 4/3-way valves with different mid-positions


A-30 Exercise 7

Exercise

Parts are fed through a drying oven on a steel chain conveyor belt. It must be possible to correct the tracking of the belt by means of a tensioning device to ensure that the belt does not run off its rollers. This device consists of a steel roller fixed at one end and movable at the other by means of a double-acting cylinder. Hydraulic power must be available continuously. The hydraulic system must switch to the recirculating (pump bypass) condition when the directional control valve is not actuated. The clamping station causes a continuous counter force to act on the cylinder. A piloted non-return valve is used to prevent creepage of the piston rod of the positioning cylinder as a result of oil leakage losses in the directional control valve. For the purposes of comparison, calculate the required drive power for circuits firstly with a 4/3-way valve, recirculating in mid-position and secondly with a 4/3-way valve, closed in mid-position.

Positional sketch


A-31 Exercise 7

EXERCISE SHEET



A-32 Exercise 7

Evaluation

Direction

Valve position

System pressure

Travel and back pressure

p0Z2

p1S1

p1S2

Advance stroke

Return stroke

Mid-position

Calculation of drive power::

PDR =

p⋅q η

Characteristic data required for calculation: PDR = Required drive power p

= System pressure supplied by pump:

Maximum 50 bar

q

= Flow rate of pump:

Constant 2 l/min

η

= Pump efficiency:

Approx. 0.7

Drive power with closed mid-position: PDR = Drive power with recirculating mid-position: PDR =

Conclusion

What is the advantage of a recirculating (bypass) circuit?


A-33 Exercise 8

Hydraulics

Subject

Cold-store door

Title

To show the use of a hydraulic accumulator as a power source To show how to use the accumulator to power advance and return

Training aim

strokes of the cylinder after the pump is switched off

Drawing the hydraulic circuit diagram Determining the necessary components Practical assembly of the circuit Determining the number of working cycles possible after the pump is switched off

Drawing conclusions Explaining the design and mode of operation of a diaphragm accumulator

Naming possible applications of an accumulator


Problem definition

A-34 Exercise 8

Exercise

A heavy cold-store door is opened and closed by a hydraulic cylinder. A hydraulic accumulator is to be installed to allow the door to be closed in the case of an electrical power failure. This will permit the cold-store door to be opened and closed a number of times. A 4/2-way valve is to be used to activate the cylinder. This valve should be connected up in such a way that the piston rod is advanced with the valve in its normal position. No provision will be made here for the safety cut-out which is essential to prevent persons from becoming trapped in the door. This cut-out function is normally provided by an electrical control device for the hydraulic system.

Be sure to follow the operating instructions for the accumulator. After switching off the control system, do not dismantle the hydraulic components until you have relieved the pressure in the accumulator and isolated this from the control system by means of the built-in shut-off valve. It is essential to relieve the accumulator pressure via a flow control valve! Positional sketch


A-35 Exercise 8

EXERCISE SHEET



A-36 Exercise 8

Evaluation

System pressure

Opening

Closing

20 bar 50 bar

Conclusion

What is the effect of fitting an accumulator to this circuit?

Explain the design and function of a diaphragm accumulator.

Name examples of applications of accumulators.


A-37 Exercise 9

Hydraulics

Subject


Title

To familiarise the student with the use of a 2-way flow control valve To show how to assemble a counter-holding circuit

Training aim

Problem definition

Understanding of a hydraulic circuit diagram Practical assembly of the circuit Commissioning a circuit with a flow control valve and counter-holding Adjustment and measurement of inlet and outlet pressures and cylinder travel time

Comparison of cylinder advance-stroke times for various inlet and outlet pressures


A-38 Exercise 9

Exercise

Several stations on a rotary machining station are driven by a hydraulic power pack. As individual stations are switched on and off, they produce pressure fluctuations throughout the hydraulic circuit. This effect will be studied on a drilling station. The fluctuations in pressure and the tractive forces created during drilling must not affect the feed of the drilling station. A flow control valve is to be used to ensure a smooth adjustable feed rate, while a pressure relief valve is to be used as a counter-holding valve to compensate for the tractive forces.

Positional sketch


A-39 Exercise 9

EXERCISE SHEET



A-40 Exercise 9

Evaluation

Measure the following: p1Z1 = Pressure upstream of flow control valve p1Z3 = Pressure downstream of flow control valve p1Z4 = Pressure at counter-holding valve t→

Fluctuating inlet pressure

Fluctuating outlet pressure

Conclusion

= Advance-stroke time of cylinder p1Z1

p1Z3

p1Z4

50 bar

10 bar

40 bar

10 bar

30 bar

10 bar

20 bar

10 bar

10 bar

10 bar

p1Z1

p1Z3

p1Z4

50 bar

10 bar

50 bar

20 bar

50 bar

30 bar

50 bar

40 bar

50 bar

50 bar

t→

t→

How does the travel change as the pressures at the inlet and outlet vary?


A-41 Exercise 10

Hydraulics

Subject

Painting booth

Title

To show how to plot a characteristic for a 2-way flow control valve To show how to make a comparison between a 2-way flow control

Training aim

valve and a throttle-type flow control valve

Drawing the hydraulic circuit diagram Practical assembly of the circuit Measurement of pressure and flow rate Plotting the characteristic of the 2-way flow control valve Comparison with a throttle valve


Problem definition

A-42 Exercise 10

Exercise

An endless chain conveyor feeds workpieces through a painting booth. The chain is driven by a hydraulic motor via a right-angle gear unit. Due to changes in the production process, the weight of the workpieces passing through the painting booth has changed. The speed of the conveyor should, however, remain the same as before. It must be determined whether this can be achieved by fitting a flow control valve, and if so which type is suitable.

Positional sketch


A-43 Exercise 10

EXERCISE SHEET

Evaluation

Measure the following: p1Z1

= Pressure upstream of valve

p1Z2

= Pressure downstream of valve

qTWFCV

= Flow rate through 2-way flow control valve

qTV

= Flow rate through throttle valve p1Z1

p1Z2

qSRV

qDV

50 bar

10 bar

2 l/min

2 l/min

50 bar

20 bar

50 bar

30 bar

50 bar

40 bar

50 bar

50 bar

p1Z1

p1Z2

qSRV

qDV

50 bar

10 bar

2 l/min

2 l/min

40 bar

10 bar

30 bar

10 bar

20 bar

10 bar

10 bar

10 bar


Fluctuating load pressure


A-44 Exercise 10

Flow control valve characteristic

Conclusion

Which valve is suitable for this application and why?


A-45 Exercise 11

Hydraulics

Subject

Embossing machine

Title

To familiarise the student with the use of a one-way flow control valve To show how to explain the difference between a flow control valve

Training aim

and throttle valve on the basis of a concrete application

Drawing the hydraulic circuit diagram Practical assembly of the circuit Commissioning a circuit with a one-way flow control valve and counter-holding

Adjustment and measurement of inlet and outlet pressures and cylinder advance-stroke time

Comparison of advance-stroke times with those in exercise 9


Problem definition

A-46 Exercise 11

Exercise

A special machine is used to emboss graphic symbols on metal foil. The foil is fed through the embossing machine with an adjustable cycle time. The downward motion of the stamp must be capable of being varied in accordance with the feed speed. The return motion must always be executed as a rapid traverse. A one-way flow control valve is used to control the speed of the stamp, while a pressure relief valve is used to prevent the weight of the stamp from pulling the piston rod out of the cylinder. A 4/2-way valve is used to switch between upwards and downwards motion.

Positional sketch


A-47 Exercise 11

EXERCISE SHEET



A-48 Exercise 11

Evaluation

Measure the following: p1Z1 = Pressure upstream of one-way flow control valve p1Z3 = Pressure downstream of one-way flow control valve p1Z4 = Pressure at counter-holding valve t→



Conclusion

= Cylinder advance-stroke time p1Z1

p1Z3

p1Z4

50 bar

10 bar

40 bar

10 bar

30 bar

10 bar

20 bar

10 bar

10 bar

10 bar

p1Z1

p1Z3

p1Z4

50 bar

10 bar

50 bar

20 bar

50 bar

30 bar

50 bar

40 bar

50 bar

50 bar

t→

t→

How does the travel time change as the pressures at the inlet and outlet vary?

What is the difference between this circuit and the one with the 2-way flow control valve (see exercise 9) and what is the reason for this?


A-49 Exercise 12

Hydraulics

Subject


Title

To familiarise the student with the design and mode of operation of a

Training aim

differential circuit

To show how to explain the influence of pressures, forces, speeds and travel times

Understanding a hydraulic circuit diagram Practical assembly of the circuit Measuring advance and return stroke times and travel and back pressures

Calculation of ratios for area and force Calculation of the flow rate through the flow control valve Comparison of this circuit with the one in exercise 6


Problem definition

A-50 Exercise 12

Exercise

The grinding table of a surface grinding machine is driven by a hydraulic cylinder. Since the speed is required to be the same in both directions, the hydraulic control circuit must be designed to provide compensation for the difference in volume of the two cylinder chambers. A differential circuit is suggested with a 3/2-way valve and a flow control valve for speed adjustment.

Positional sketch


A-51 Exercise 12

EXERCISE SHEET



A-52 Exercise 12

Evaluation

Measure the following: p1Z1 = Pressure on piston side of cylinder p1Z2 = Pressure on annular side of cylinder p0Z2 = System pressure = 50 bar t→

Values table

= Cylinder advance-stroke time approx. 4 s Direction

p1Z1

p1Z2

t

Advance stroke Return stroke

Cylinder dimensions: 2

Piston area:

APN = 2.0 cm


APR = 1.2 cm

2

Cylinder stroke:

s = 0.2 m A PN = A PR

Area ratio:

α=

Time ratio:

t adv = t ret

Force ratio:

F1 A PN ⋅ p1Z1 = = F2 A PR ⋅ p1Z 2

Flow rate during advance stroke: Piston side:

qPN = A PN ⋅

Piston annular side:

qPR = A PR ⋅

s t adv s t adv

=

=


A-53 Exercise 12

EXERCISE SHEET

Flow control valve component: qFCV = qPN − qPR =

Flow rate during return stroke: Piston annular side:

qPR = A PR ⋅

s t ret

=

When the 3/2-way valve is activated, the same pressure is present at both cylinder ports. Why does the piston advance?

During the advance stroke, the pressures in the two cylinder chambers are different. Why does the piston advance despite the fact that the travel pressure is lower than the back pressure?

What force can the cylinder exert during its advance stroke?


Conclusion

A-54 Exercise 12

What is the difference between this differential circuit and a simple cylinder control circuit (one connection to each of P and T as, for example, in exercise 6)? 1. What are the factors governing the advance-stroke speed vadv? 2. What is the value of the return-stroke speed vret in comparison with the advance-stroke speed vadv? 3. What are the factors governing the advance-stroke time tadv? 4. What is the value of the return-stroke time tret in comparison with the advance-stroke time tadv? Vergleich

System

Simple cylinder control circuit

Differential circuit

1. Advance-stroke speed vadv 2. Return-stroke speed vret 3. Advance-stroke time tadv 4. Return-stroke time tret

Conclusion

What area ratio results in identical advance and return stroke speeds (using a differential circuit)?


A-55 Exercise 13

Hydraulics

Subject

Drilling machine

Title

To teach the student how to design a control circuit with reduced out-

Training aim

put pressure

To show how to explain the mode of operation of a 3-way pressure regulator

Drawing the hydraulic circuit diagram Practical assembly of the circuit Measuring the travel and back pressures Setting a counter pressure Assessment of the effect of using a pressure regulator


Problem definition

A-56 Exercise 13

Exercise

A drilling machine is used for work on various hollow workpieces. The workpieces are hydraulically clamped in a vice. It must be possible to reduce the clamping pressure to suit the design of the workpiece. It must also be possible to vary the closing speed by means of a one-way flow control valve.

Positional sketch

1A


A-57 Exercise 13

EXERCISE SHEET

Evaluation

Measure the following: p1Z1 = Pressure upstream of flow control valve p1Z2 = Pressure upstream of cylinder p1Z3 = Pressure downstream of cylinder Study the following cases: 1. Piston advance stroke 2. Piston advanced to end position with setting p1Z2 = 15 bar. 3. Piston advance stroke with counter pressure setting, p1Z3 = 20 bar. 4. Piston advanced to end position 5. Piston advance stroke with shut-off valve closed 6. Piston advanced to end position with shut-off valve closed Cases of examination

p1Z1

p1Z2

p1Z3

Advance stroke

p1Z1

p1Z2

p1Z3

Return stroke

1. Advance stroke 2. End position 3. Advance stroke with counter pressure 4. End position 5. Advance stroke with pressure regulator 6. End position

Cases of examination 1. Return stroke 2. End position 3. Return stroke with counter pressure 4. End position 5. Return stroke with pressure regulator 6. End position


A-58 Exercise 13

Conclusion

When is it appropriate to use a pressure regulator?

What possible disadvantage may result from the use of a pressure regulator?


A-59 Exercise 14

Hydraulics

Subject

Bulkhead door

Title

To familiarise the student with a circuit for the hydraulic clamping of a

Training aim

bulkhead door

To demonstrate a comparison of circuits with and without counterholding

Drawing the hydraulic circuit diagram Practical assembly of the circuit Measuring the cylinder advance-stroke time with and without a load and with and without counter-holding

Comparison and assessment of results


Problem definition

A-60 Exercise 14

Exercise

A double-acting cylinder is used to open and close a bulkhead door. Closing must be carried out smoothly and at a constant adjustable speed. The speed is adjusted by means of a one-way flow control valve. A pressure relief valve must be fitted to provide counter-holding and prevent the heavy door from pulling the piston rod out of the cylinder during the closing operation.

Positional sketch


A-61 Exercise 14

EXERCISE SHEET



A-62 Exercise 14

Evaluation

Measure the following: t→

= Cylinder advance-stroke time

p1Z1 = Cylinder travel pressure p1Z2 = Cylinder back pressure p0Z2 = System pressure The applied load and counter-holding should now be varied. Initial settings should be such as to achieve an advance-stroke time of 5 s with a system pressure of 50 bar but without an applied load or counterholding. 10 bar back pressure should subsequently be set.

When dismantling the circuit, ensure that no pressurised fluid is trapped (p1Z2 = 0 bar). Values table

Load and counter-holding Without load or counter-holding

p0Z2

p1Z1

p1Z2

t→ 5s

50 bar

With load without counter-holding

Conclusion

With load and counter-holding

10 bar

Without load with counter-holding

10 bar

How does the travel time vary as the load changes?

Which circuit is more suitable?


A-63 Exercise 15

Hydraulics

Subject

Ferry loading ramp

Title

To familiarise the student with a speed control circuit with a tractive

Training aim

load

To compare circuits with flow control valves in the inlet line and outlet line respectively

Drawing the hydraulic circuit diagram Practical assembly of the circuit Measuring the cylinder advance time and travel and back pressures with flow control valves in the inlet line and outlet line respectively

Comparison and assessment of results


Problem definition

A-64 Exercise 15

Exercise

The loading ramp of a car ferry must be capable of being set to different heights. The ramp is raised and lowered by a hydraulic cylinder. This motion must be carried out smoothly and at a constant speed. A flow control valve is to be used to adjust the speed. This must be installed in such a way as to prevent excessive pressures from developing within the system.

Positional sketch


A-65 Exercise 15

EXERCISE SHEET



A-66 Exercise 15

Evaluation



p1Z1 = Cylinder travel pressure p1Z2 = Cylinder back pressure p0Z2 = System pressure Vary the following:

Applied load Counter-holding Flow control in inlet and outlet lines Settings:

First, without an applied load or counter-holding and with a flow control valve in the inlet line, make settings to obtain an advance-stroke time of t –> = 5 s with a system pressure of p0Z2 = 50 bar.

Then set a counter pressure of p1Z2 = 10 bar. Then use a flow control valve in the outlet line to provide counterholding.

When dismantling the circuit, ensure that no pressurised fluid is trapped (p1Z2 = 0 bar).


A-67 Exercise 15

EXERCISE SHEET

Load and counter-holding Without load or counter-holding

p0Z2

p1Z1

p1Z2

t→

Flow control valve in inlet line

5s

50 bar

With load without counter-holding With load and counter-holding

10 bar


10 bar

Load Without load

p0Z2

p1Z1

50 bar

p1Z2

t→

Flow control valve in outlet line

5s

With load

How does the travel time change as the load is varied?

Which circuit is more suitable?


Conclusion

A-68 Exercise 15


A-69 Exercise 16

Hydraulics

Subject

Skip handling

Title

To develop a hydraulic circuit for a double-acting cylinder subject to a

Training aim

varying load

Drawing the circuit diagram Practical assembly of the circuit Commissioning of control circuit Description of mode of operation of control circuit


Problem definition

A-70 Exercise 16

Exercise

The loading and unloading of skips from a skip transporter is carried out using two double-acting cylinders. Each cylinder is subject to varying loads – tractive load during unloading and compressive load during loading. The skip should be raised and lowered at a slow constant speed. Each cylinder must therefore be hydraulically clamped on both sides.

Positional sketch


A-71 Exercise 16

EXERCISE SHEET



A-72 Exercise 16

Conclusion

How is hydraulic clamping produced on both sides?


A-73 Exercise 17

Hydraulics

Subject

Bonding press

Title

To teach the student how to specify the pressure for a double-acting

Training aim

cylinder

To show how to choose either a pressure relief valve or a pressure regulator

Drawing the hydraulic circuit diagram Practical assembly of the circuit Measurement and comparison of system pressure, travel pressure and final pressure

Assessment of the suitability of a pressure relief valve and pressure regulator


Problem definition

A-74 Exercise 17

Exercise

A bonding press is used to stick pictures or lettering onto wood or plastic panels. The working pressure must be adjustable to suit the base material and adhesive used and must be capable of being maintained for a long time while the directional control valve is activated. Develop and compare two circuits. The first should use a 3-way pressure regulator to adjust the press pressure, while the second should incorporate a pressure relief valve connected into the bypass line for this purpose. A 4/3-way valve should be used for activation in both cases.

Positional sketch


A-75 Exercise 17

EXERCISE SHEET



A-76 Exercise 17

Evaluation

Conclusion

Carry out the following settings: p0Z2 = System pressure

= 50 bar

p1Z2 = Pressure upstream of cylinder

= 30 bar

With which circuit does the system pressure vary as the cylinder advances?

When is it advantageous to use the pressure relief valve?


A-77 Exercise 18

Hydraulics

Subject

Assembly device

Title

To familiarise the student with a pressure sequence circuit To teach the student how to draw a displacement-step diagram

Training aim

Problem definition

Development of hydraulic circuit diagram Drawing the displacement-step diagram Practical assembly of the circuit Systematic commissioning with setting of pressure and flow rate


A-78 Exercise 18

Exercise

An assembly device is used to press workpieces together for drilling. Cylinder 1A1 presses a workpiece into the housing. This operation should be carried out slowly at a constant speed. When the pressure in cylinder 1A1 has reached 20 bar (workpiece pressed into place), a hole is drilled through the workpiece and housing. The drill is driven by a hydraulic motor. After the drilling operation, the drill is switched off and retracted (1A2). Cylinder 1A1 is retracted only when the drill has withdrawn from the housing.

Positional sketch

1A1

1A2


A-79 Exercise 18

EXERCISE SHEET



A-80 Exercise 18

Displacement-step diagram

Time

Components

Step Description

Conclusion

Designa tion

Signal

What are the points to note when commissioning the control circuit? 1. 2. 3. 4. 5. 6.


A-81 Exercise 19

Hydraulics

Subject

Calculations for an assembly device

Title

To enable the student to calculate the forces associated witha dou-

Training aim

ble-acting cylinder

To enable the student to calculate the advance-stroke time of a cylinder piston

Writing a problem description Calculating the press-fitting force Calculating the press-fitting time


Problem definition

A-82 Exercise 19

Exercise

An assembly device is used to press workpieces together for drilling. The operating sequence is described in exercise 18. Our objective here is to check the pressing operation of cylinder 1A1 mathematically. Determine the press-fitting force using the given data. Note that, while the press-fitting pressure is available as specified, the resistances of the lines and directional control valve cause an opposing pressure to act on the annular piston side, thus reducing the actual force available.The flow rate is kept constant by a flow control valve. This together with the cylinder stroke is used to calculate the travel time for the press-fitting operation.

Positional sketch

1A1


A-83 Exercise 19

EXERCISE SHEET

Characteristic data of control system:

Evaluation

Cylinder: Piston diameter

D = 50 mm

Piston rod diamete

d = 25 mm

Stroke

s = 250 mm

Hydraulic system: Flow rate

q = 5 l/min

Press-fitting pressure

p1 = 50 bar

Counter pressure

p2 = 6 bar Schematic diagram


A-84 Exercise 19

Piston force: F1 = A PN ⋅ p1 =

Counter force: F2 = A PR ⋅ p 2 =

Press-fitting force: F = F1 − F2 =

Press-fitting time: t=

V = q


A-85 Exercise 20

Hydraulics

Subject

Tipping container

Title

To familiarise the student with an electrohydraulic circuit

Training aim

Development of hydraulic and electrical circuit diagrams Assembly of control system

Problem definition


A-86 Exercise 20

Exercise

A conveyor belt transports metal swarf into a tipping container. When the container is full, it is emptied into a truck. A double-acting cylinder is used for this purpose, activated by a solenoid-actuated 4/3-way valve. The piston rod of the cylinder is advanced while the container is in position to receive swarf. To enable the hydraulic power pack to be switched off during this time, the piston rod of the cylinder must be protected by hydraulic means against undesired retraction (caused by leakage in the valve). The electrical activation of the valve must be manually controlled, i.e. the cylinder must move only when the “Up” or “Down” pushbuttons are pressed.

Positional sketch


A-87 Exercise 20

EXERCISE SHEET



A-88 Exercise 20

Circuit diagram,electrical

S1 = “Up” pushbutton S2 = “Down” pushbutton

Conclusion

What measure ensures that the cylinder maintains its position and does not move even if the “Up” and “Down” pushbuttons are accidentally pressed simultaneously?


B-1 Fundamentals

Section B – Fundamentals The theoretical fundamentals for the “Hydraulics” training package are summarised in the textbook:

Hydraulics Basic Level TP501


B-2 Fundamentals


C-1

Section C – Solutions Solution 1:

Automatic lathe

C-3

Solution 2:


C-7

Solution 3:

Drawing press

C-11

Solution 4:


C-15

Solution 5:

Hardening furnace

C-19

Solution 6:


C-23

Solution 7:


C-27

Solution 8:

Cold-store door

C-33

Solution 9:


C-37

Solution 10:

Painting booth

C-41

Solution 11:

Embossing machine

C-45

Solution 12:


C-49

Solution 13:

Drilling machine

C-59

Solution 14:

Bulkhead door

C-65

Solution 15:

Ferry loading ramp

C-69

Solution 16:

Skip handling

C-73

Solution 17:

Bonding press

C-77

Solution 18:

Assembly device

C-79

Solution 19:


C-83

Solution 20:

Tipping container

C-85


C-2


C-3 Solution 1

Automatic lathe


Practical assembly, hydraulic


C-4 Solution 1

Components list

Solution description

Evaluation

Item no.

Qty.

Description

0Z1

1


0Z2

1

Pressure gauge

1V

1

Shut-off valve

1S

1

Flow sensor

3

Hose line

Once the hydraulic circuit has been assembled, valve 1V should be fully opened. Now close this valve slowly to set the first p value as shown on the pressure gauge 0Z2. The maximum attainable pressure is 60 bar, governed by a pressure relief valve built into the pump which is set to this value.

System pressure p Flow rate q

15

20

25

30

35

40

45

50

2.33 2.31 2.29 2.28 2.26 2.24 2.22 2.20

bar l/min

Pump characteristic


C-5 Solution 1

As the pressure rises, the pump delivery falls slightly. In theory, the characteristic curve for the pump should be a straight line.The decrease in pump delivery is due to internal leakage losses, which become greater as the pressure increases The ratio of the measured pump delivery and theoretical pump delivery is the effective volumetric efficiency of the pump.

For technical reasons, the actual value recorded in this exercise is the power consumption of the electric motor or the premature opening of the pressure relief valve. The pump is dimensioned for a maximum pressure of 250 bar (see data sheet). An electric motor with an appropriately high rating would be required to achieve this. This would not, however, be meaningful, since the exercises are carried out with a maximum pressure of 60 bar.


Conclusions

C-6 Solution 1


C-7 Solution 2




C-8 Solution 2


System pressure p = 50 bar (5 MPa) (Shut-off valve 1V1 closed) Measured value q in l/min Pump safety valve pmax = 60 bar (6 MPa)

Components list

Item no.

Qty.

Description

0Z1

1


0Z2

1

Pressure gauge

1V1

1

Shut-off valve

1V2

1


1S

1

Flow sensor

5

Hose line

2

Branch tee


C-9 Solution 2

Once the hydraulic circuit has been assembled and checked, valve 1V1 should be closed and the pressure relief valve 1V2 fully opened. Switch on the hydraulic power pack and close the pressure relief valve until the pressure gauge 0Z2 indicates 50 bar. Now fully open shut-off valve 1V1 and close it again in steps to set the pressures specified in the table; evaluate the associated flow rates. At the same time, observe the pressure at which the valve begins to open.


If, at 50 bar pressure, a flow rate of 2.3 l/min is not measured at the pressure relief valve, this indicates that the pressure relief valve fitted directly to the pump is already starting to open.

Remark

Working pressure p Flow rate q

35

40

42.5

45

47.5

50

bar

0

0

0

0.2

1.17

2.15

l/min

Evaluation

Pressure relief valve characteristic Maximum pressure

Opening pressure


C-10 Solution 2

Conclusions

Every pressure relief valve has a certain opening pressure at which point diversion of the flow through the valve begins. The difference between opening pressure and maximum pressure is 5 bar in this case. When the preset maximum pressure is reached, the entire pump delivery is discharged via the pressure relief valve.

A piloted pressure relief valve can also be used to record the characteristic. Due to the low flow rate, the same shape of characteristic curve will be obtained.


C-11 Solution 3

Drawing press



C-12 Solution 3


System pressure p = 50 bar (5 MPa)


C-13 Solution 3

Item no.

Qty.

Description

0Z1

1


0Z2, 0Z3, 0Z4

3

Pressure gauge

0V1

1


0V2

1

Flow control valve

0V3

1

Shut-off valve

1V4

1


1V3

1

Throttle valve

1V2

1

4/2-way valve

1V1

1

4/3-way valve

1S

1

Flow sensor

7

Hose line

3

Branch tee

Once the hydraulic circuit has been assembled and checked, the shutoff valve 0V3 should be closed and the pressure relief valve 0V1 fully opened. Switch on the hydraulic power pack and close the pressure relief valve until the pressure gauge 0V1 indicates 50 bar. Now carry out the series of measurements specified in the table. Adjust the flow rate by means of the flow control valve 0V2 and measure it with the flow sensor 1S.

Pressure sensors are recommended for the measurement of pressures at items 0Z3 and 0Z4.


Components list


C-14 Solution 3

Evaluation

p0Z3 = Pressure upstream of component p0Z4 = Pressure downstream of component

Values table

Component

Pressure relief valve, fully open Throttle valve, fully open 4/2-way valve, P –> A 4/3-way valve, P –> A

Conclusions

Flow rate q l/min

Pressure p0Z3 bar

Pressure p0Z4 bar

Pressure difference ∆p bar

2

4.6

2.5

2.1

1

1.9

1.0

0.9

2

4.3

2.5

1.8

1

1.9

1.2

0.7

2

4.0

2.5

1.5

1

1.9

1.2

0.7

2

4.3

2.5

1.8

1

1.8

1.1

0.7

When the flow rate doubles, the pressure difference increases by even more than this. The hydraulic resistance increases. This pressure loss means a loss of power.


C-15 Solution 4




C-16 Solution 4



Pump safety valve pmax = 60 bar (6 MPa)


C-17 Solution 4

Item no.

Qty.

Description

0Z1

1


0V1

1

Non-return valve (5 bar)

0Z2

1

Pressure gauge

0V2

1


1V

1

Shut-off valve

1A

1


1Z

1

Loading weight

8

Hose line

4

Branch tee

Components list

For this exercise, the cylinder is bolted onto the base plate on the left of the profile plate and loaded with the weight. When the cylinder is connected up, it is essential that the upper connection is connected to the tank. Once the circuit has been assembled, the PRV 0V2 should first be fully opened. The hydraulic power pack should then be switched on and the PRV 0V2 slowly closed. The piston rod will then travel to its upper end position. Continue to close the PRV until the pressure gauge 0Z2 indicates 50bar. Now switch off the hydraulic power pack. It can be demonstrated by briefly opening the shut-off valve that the non-return valve prevents the weight from lowering further and that return flow of hydraulic fluid during the return stroke can take place only via the 2/2way valve 1V.


The piston rod can retract only when the pump is switched off. This is arranged intentionally in systems like the one shown here. This ensures that the hydraulic power pack is switched off during lengthy standstill periods.

Conclusions


C-18 Solution 4


C-19 Solution 5

Hardening furnace



C-20 Solution 5





C-21 Solution 5

Item no.

Qty.

Description

Components list

0Z1

1


0Z2, 0Z3, 1Z1

3

Pressure gauge

0V1

1

Non-return valve

0V2

1


1V

1

4/2-way valve, manually operated

1A

1


1Z2

1

Loading weight

7

Hose line

3

Branch tee

1

Stop-watch

For this exercise, the cylinder is bolted onto the base plate on the left of the profile plate and loaded with the weight. When the cylinder is connected up, it is essential that the upper connection is connected to the tank. In place of a 3/2-way valve, a 4/2-way valve is now used, with one connection blanked off. Once the circuit has been assembled, the PRV 0V2 should first be fully opened. The hydraulic power pack should then be switched on and the PRV 0V2 slowly closed until the pressure gauge 0Z3 indicates 50 bar. The 4/2-way valve 1V can now be slowly reversed, which will cause the piston rod of the cylinder to advance. The design of the valve means that, as this is slowly reversed, the full cross-section of the valve is not immediately opened. Initially, the pump delivery to the cylinder will be throttled. As soon as the valve is returned to its initial position, the piston rod of the cylinder will return to its lower end position.


The values specified in the tables can now be measured. Direction

Travel pressure

Travel time

Advance stroke

8 bar

1.1 s

Return stroke

0 bar

1.4 s


Evaluation

C-22 Solution 5

Characteristic data required for calculation: Applied load:: Piston area:

FW =

90 N

APN =

2 cm

2

Stroke length:

s = 200 mm

Pump delivery:

q=

Load pressure:

pL =

2 l/min FW 90 N 45 N = = = 4.5 bar 2 A PN 2 cm cm 2

Hydraulic resistance = Travel pressure - load pressure pres = 8 bar - 4.5 bar = 3.5 bar

Conclusions

The back pressure is considerably lower than the hydraulic resistance. A cylinder motion can take place only if this case applies. The value of the back pressure depends on the hydraulic resistances. These are very low when fluid is discharged into the tank. 2000 cm 3 l 60 s q = = min2 = A PN 2 cm 2 cm 2 2

Advance-stroke speed:

v adv

v adv = 16.67


Conclusions

t adv =

s v adv

cm m = 0.17 s s

=

0.2 m = 12 . s m 0.17 s

The measured advance-stroke time, 1.1 s., is slightly less than the calculated time. The reason for this may be that the delivery of a new pump is somewhat greater than 2 l/min.


C-23 Solution 6




C-24 Solution 6




Components list

Item no.

Qty.

Description

0Z1

1


0Z2

1

Pressure gauge

1S1, 1S2

2

Pressure sensor

0V

1


1V

1


1A

1

Cylinder

6

Hose line

2

Branch tee

1

Stop-watch


C-25 Solution 6

Once the circuit has been assembled and checked, the hydraulic power pack should be switched on and the system pressure set on the pressure relief valve 0V to 50 bar. Pressure sensors should be used to measure the travel and back pressures. Pressure gauges are sluggish in operation and would give incorrect readings.


When the hand lever of the 4/2-way valve is actuated, the piston rod of the cylinder will advance until the lever is released or the piston rod runs against the stop. When the lever is released, the piston rod will immediately return to its retracted end position. Before the pressures and times are measured, the piston rod should be advanced and retracted several times to expel any air which may have entered the piston-rod chamber during the previous exercises. Advance stroke

Return stroke


Back pressure p1S2

Travel time tadv

2.4 bar

2 bar

1.2 s

Back pressure p1S1


Travel time tein

5.3 bar

11 bar

0.8 s

Characteristic data required for calculation: Piston area:

APN =

2.0 cm

2


APR =

1.2 cm

2

Stroke length:

s = 200 mm

Pump output:

q =

Area ratio:

α=

2 l/min

A PN 2 cm 2 = = 1667 . A PR 12 . cm 2

2000 cm 3 l 60 s q = = min2 = A PN 2 cm 2 cm 2 2

Advance-stroke speed: v adv

Vadv = 16.67


cm m = 0.17 s s

Evaluation

C-26 Solution 6


Return-stroke speed:

t adv =

v ret

s v adv

=

2000 cm 3 l 2 60 s q min = = = 2 A PR 12 12 . cm . cm 2

v ret = 27.78

Conclusions

2m = 12 . s m 0.17 s

cm m = 0.28 s s

s 0.2 m = = 0.7 s m v ret 0.28 s

Return-stroke time:

t ret =

Travel speed ratio:

m 0.17 Vadv s = 0.6 = m Vret 0.28 s

Travel time ratio:

t adv 12 . s = = 17 . t ret 0.7 s

The travel speed ratio is equal to the area ratio α of the cylinder. The speed ratio is equal to the reciprocal of the area ratio.


C-27 Solution 7




C-28 Solution 7





C-29 Solution 7

Item no.

Qty.

Description

0Z1

1


0Z2

1

Pressure gauge

1S1, 1S2

2

Pressure sensor

0V1

1


0V2

1

Shut-off valve

1V1

1

4/3-way valve, manually operated, Recirculating mid-position

1V2

1

Pilot-operated non-return valve

1A

1

Cylinder

9

Hose line

3

Branch tee

1

Stop-watch

After the circuit has been assembled and tested, the shut-off valve 0V2 should be closed and the pressure relief valve 0V1 opened. Switch on the hydraulic power pack and close the PRV 0V1 until the pressure gauge 0Z1 indicates 50 bar. The shut-off valve 0V2 can now be opened. Observe when doing this that the pressure gauge 0Z1 shows an immediate drop from the set pressure of 50 bar to approx. 3 bar, since in its mid-position the 4/3-way valve 1V1 discharges the flow of hydraulic fluid to the tank. The piston rod can be brought into any desired position by actuating the 4/3-way valve. When this valve is brought into its mid-position, the piston rod immediately halts. The non-return valve prevents the piston rod from being pushed back by a counter force.


Components list


C-30 Solution 7

In conjunction with the pilot-operated non-return valve, a 4/3-way valve with a mid-position “A and B connected to T” and “P closed” should be used in order to depressurise the pilot line and supply line to the piloted non-return valve. The non-return valve can close reliably only when depressurised. The 4/3-way valve with recirculating mid-position, included in the equipment set, can also be used for these exercises. The internal leakage losses resulting from the design of this valve will also cause the nonreturn valve to close. Evaluation

Direction

Valve position

System pressure

Travel and back pressure

p0Z2

p1S1

p1S2

Advance stroke

8 bar

2.2 bar

1.6 bar

Return stroke

2.2 bar

9.4 bar

17.9 bar

Mid-position

3.1 bar

1.6 bar

1.7 bar

Calculation of drive power::

PDR =

p⋅q η

Characteristic data required for calculation: PDR = Required drive power p

= System pressure supplied by pump:

Maximum 50 bar

q

= Flow rate of pump:

Constant 2 l/min

η

= Pump efficiency:

Approx. 0.7


C-31 Solution 7

Drive power with closed mid-position:

PDR =

PDR =

l 3 3 min = 50 kp ⋅ 2 dm = 50 ⋅ 10 N ⋅ 2 ⋅ 1000 cm 0.7 0.7 cm ⋅ 60 s 0.7 cm 2 ⋅ 60 s

50 bar ⋅ 2

Ncm 3 Nm 50 ⋅ 2 50 ⋅ 2 ⋅ 10000 = ⋅ 100 = 238 W 2 s 0.7 ⋅ 60 cm ⋅ s 0.7 ⋅ 60

Drive power with bypass to pump:

PDR =

l min = 3.1⋅ 2 ⋅ 100 Nm = 15 W s 0.7 0.7 ⋅ 60

3.1 bar ⋅ 2

The 4/3-way valve with recirculating mid-position is mainly used in cases where a cylinder or motor is driven by a constant-displacement pump. In the recirculating mid-position, hydraulic fluid is discharged to the tank at almost zero pressure, which means that the temperature rise remains small. The disadvantage of using this valve is that it is not possible to operate any further hydraulic circuits. In the case of valves with a closed position for connection P, the pump delivery is discharged to the tank at maximum system pressure, which results in pronounced heating of the fluid (= energy loss).


Conclusions

C-32 Solution 7


C-33 Solution 8

Cold-store door



C-34 Solution 8





C-35 Solution 8

Item no.

Qty.

Description

Components list

0Z1

1


0Z2, 0Z3

2

Pressure gauge

0V1

1


0V2

1

Non-return valve

0V3

1

One-way flow -return valve, adjustable

0Z4

1


1V

1


1A

1

double-acting Cylinder

7

Hose line

3

Branch tee

After the circuit has been assembled and checked, the accumulator should first be switched off and the pressure relief valve 0V1 fully opened. Now switch on the hydraulic power pack and set the system pressure to 50 bar. The accumulator can now be charged. Allow the cylinder to advance and retract several times and then switch off the hydraulic power pack. It is possible to advance and retract the cylinder several times more by actuating the 4/2-way valve 1A. Following this, the accumulator pressure will fall slowly, as indicated by the pressure gauge 0V3. Be sure to switch off and depressurise the accumulator before dismantling the circuit!

System pressure

Opening

Closing

20 bar

2x

1x

50 bar

4x

3x



Evaluation

C-36 Solution 8

Conclusions

Without the accumulator fitted, the door will remain in its instantaneous position after a power failure and it will no longer be possible to move it. This diaphragm accumulator allows the door to be opened 2 x and closed 1 x with a system pressure of 20 bar and opened 4 x and closed 3 x with a system pressure of 50 bar. The higher the hydraulic pressure with which the accumulator is charged, the more times the door can be opened and closed.

Design

In the case of a diaphragm accumulator, a diaphragm is clamped into place in the pressure vessel to act as a divider between the hydraulic fluid and the gas cushion (nitrogen). A gas valve is fitted at the top to allow the accumulator to be pressurised with gas via a filling device. A closure head is fitted to the diaphragm or a shut-off valve to the gas inlet to prevent the diaphragm from creeping into the gas inlet as the gas is discharged and becoming damaged. The accumulator used here has an 3 initial gas pressure rating of 10 bar and a nominal volume of 0.32 cm . All accumulators must be fitted with a lead-sealed safety pressure relief valve and a shut-off valve, as appropriate to their capacity.

Mode of operation

When hydraulic fluid is forced into the accumulator, this causes a corresponding reduction in the volume of the gas. At the same time, the pressure in the gas cushion rises until the gas and hydraulic fluid are at the same pressure. When the fluid pressure falls, the gas forces fluid back into the hydraulic system. A non-return valve must be fitted upstream of the pump to prevent stored fluid from being discharged via the pump when this is switched off. Comprehensive manufacturers’ tables are available for use when sizing accumulators.In addition to diaphragm and bladder accumulators, piston accumulators are also available if large capacities are required.

Examples of applications

Accumulators are used for the following:

Compensation for leakage losses Energy reserve in emergencies Compensation for peaks in flow rate demand Cushioning of switching jolts


C-37 Solution 9






C-38 Solution 9

Components list


Item no.

Qty.

Description

0Z1

1


0Z2, 1Z1, 1Z2, 1Z3, 1Z4

5

Pressure gauge

0V1, 1V4

2


0V2

1

Shut-off valve

1V1

1


1V3

1

2-way flow control valve

1V2, 1V5

2

Non-return valve

1A

1

Cylinder

12

Hose line

6

Branch tee

1

Stop-watch

Assemble and check the circuit. Close the shut-off valve 0V2 and set the desired pressure by means of the pressure relief valve 0V1. Now open the pressure relief valve 1V4 and the shut-off valve 0V2. Open the flow control valve approx. 2 turns so that the piston rod moves into its forward end position in approx. 5 sec., when the 4/3-way valve is actuated. Do not make any further changes to the flow control valve setting. As soon as the piston rod reaches the forward end position with the 4/3-way valve actuated, use the pressure relief valve 0V1 to set the values in table 1 (check these on the pressure gauge 1Z1). The pressure as indicated on the pressure gauge 1Z4 must be set during the advance stroke, using the pressure relief valve 1V4. Flow is not possible through the flow control valve and pressure relief valve in the opposite direction. The two non-return valves 1V2 and 1V5 are fitted to allow these to be bypassed.

In the case of settings p1Z1 = 50 bar and p1Z4 = 40 bar (in tables 1 and 2), the pump requires approx. 1-2 sec. to build up a counter pressure of 40 bar. The time should therefore be measured not from the moment the 4/3-way valve is actuated but from the moment the piston rod starts to move. In the case of table 2, the specified values of 50 bar cannot be fully reached due to the resistances present.


C-39 Solution 9

Evaluation

p1Z1 = Pressure upstream of flow control valve p1Z3 = Pressure downstream of flow control valve p1Z4 = Pressure at counter-holding valve t→


p1Z3

p1Z4

t→

50 bar

7 bar

10 bar

4s

40 bar

7 bar

10 bar

4s

30 bar

7 bar

10 bar

4s

20 bar

7 bar

10 bar

4s

10 bar

7 bar

10 bar

6s

p1Z1

p1Z3

p1Z4

t→

50 bar

7 bar

10 bar

4s

50 bar

13 bar

20 bar

4s

50 bar

20 bar

30 bar

4s

50 bar

26 bar

40 bar

4s

50 bar

32 bar

50 bar

4s

Even with modified pressures at the inlet and outlet, the piston advancestroke times remain constant. The flow rate will be inadequate only if the supply pressure is too low. The pressure intensification effect becomes noticeable with higher counter pressure; the advance-stroke speed falls only when the counter pressure reaches approx. 70 bar. The pressure reached on the piston rod side is then 48 bar.

This exercise is also suitable for practice with fault-finding. If the nonreturn valves are installed incorrectly, the piston rod will not retract. The cause of this can be identified by systematic observation of the pressure-gauge readings.




Conclusions

C-40 Solution 9


C-41 Solution 10

Painting booth



C-42 Solution 10




Components list

Item no.

Qty.

Description

0Z1

1


0Z2, 1Z1, 1Z2

3

Pressure gauge

0V1, 1V1

2


0V2

1

Shut-off valve

1S

1

Flow sensor

1V2

1

Flow control valve

1V3

2

Throttle valve

6

Hose line

2

Branch tee


C-43 Solution 10

Assemble and check the circuit in accordance with the circuit diagram. Fully open the pressure relief valves 0V1 and 1V1 and close the shut-off valve 0V2. Open the flow control valve 0Z2 approx. 2 turns.The hydraulic power pack can now be switched on. The system pressure of 50 bar required for the exercise should be set on the pressure relief valve 0V1 and checked on the pressure gauge 0Z2.


Now open the shut-off valve 0V2. If the pressure gauge 1Z1 shows less than 50 bar, re-adjust the pressure relief valve 0V1 slightly. The flow control valve 0V2 can now be set to the desired flow rate of 2 l/min. The load pressure should be varied by means of the pressure relief valve 1V1 in accordance with the specified values. For the second half of the measurements, the pressure relief valve 1V1 should be fully opened and the system pressure varied by means of the pressure relief valve 0V1. The system pressure/flow rate characteristic for the flow control valve can then be plotted. If the same exercise is carried out using a throttle-type flow control valve, the differences will be readily apparent in the tables of values.

Evaluation

p1Z1 = Pressure upstream of valve p1Z2 = Pressure downstream of valve qFCV = Flow rate through flow control valve qTV

= Flow rate through throttle valve p1Z1

p1Z2

qSRV

qDV

50 bar

10 bar

2 l/min

2 l/min

50 bar

20 bar

2 l/min

1.8 l/min

50 bar

30 bar

2 l/min

1.3 l/min

50 bar

40 bar

2 l/min

0.7 l/min

50 bar

50 bar

1.2 l/min

0.1 l/min


Fluctuating load pressure

C-44 Solution 10


p1Z1

p1Z2

qFCV

qTV

50 bar

10 bar

2 l/min

2 l/min

40 bar

10 bar

2 l/min

1.6 l/min

30 bar

10 bar

2 l/min

1.3 l/min

20 bar

10 bar

2 l/min

0.8 l/min

10 bar

10 bar

1.5 l/min

0.4 l/min

Flow control valve characteristic

Conclusions

Only the flow control valve offers a suitable means of setting a constant speed with different pressures. In the case of the throttle valve, the flow rate varies as a function of pressure. Reason: In the case of the flow control valve, the built-in pressure compensator keeps the pressure difference constant. This gives a constant flow rate, which can then be adjusted with a throttle valve. Operation of the pressure compensator does, however, require a certain minimum pressure. The throttle valve is a simple restrictor, which produces a flow rate as a function of the pressure difference.


C-45 Solution 11

Embossing machine





C-46 Solution 11

Components list


Item no.

Qty.

Description

0Z1

1


0Z2, 1Z1, 1Z2, 1Z3, 1Z4

5

Pressure gauge

0V1, 1V3

2


0V2

1

Shut-off valve

1V1

1


1V2

1


1A

1


1V2

1

Non-return valve

14

Hose line

4

Branch tee

1

Stop-watch

Once the circuit has been assembled and checked, close the shut-off valve 0V2 and set a pressure of 50 bar, using the pressure relief valve 0V1. Open the pressure relief valve 1V3 and the shut-off valve. Now adjust the one-way flow control valve 1V2 in such a way that the piston rod reaches its forward end position in approx. 5 sec. after the 4/2-way valve 1V1 is reversed. Do not make any further changes to the setting of the one-way flow control valve. The pressure specified in table 1 of 10 bar, as indicated by the pressure gauge 1Z4, can be set only during the advance stroke, using the pressure relief valve 0V1. The pressure p1Z1 should be set by means of the pressure relief valve 0V1 as soon as the 4/2-way valve is reversed and the piston rod has reached its forward end position.


C-47 Solution 11

Evaluation

p1Z1 = Pressure upstream of one-way flow control valve p1Z3 = Pressure downstream of one-way flow control valve p1Z4 = Pressure at counter-holding valve t→


p1Z3

p1Z4

t→

50 bar

9 bar

10 bar

4s

40 bar

9 bar

10 bar

5s

30 bar

9 bar

10 bar

7.5 s

20 bar

9.5 bar

10 bar

12.5 s

10 bar

9.8 bar

10 bar

57 s

p1Z1

p1Z3

p1Z4

t→

50 bar

9 bar

10 bar

4s

50 bar

15 bar

20 bar

4.5 s

50 bar

22 bar

30 bar

5s

50 bar

28 bar

40 bar

6.5 s

50 bar

35 bar

50 bar

7s




C-48 Solution 11

Conclusions

In the case of a circuit with a throttle valve, the travel speed falls both as the inlet pressure is reduced and as the counter pressure increases. In the case of the circuit with a flow control valve (exercise 9), the travel speed remains constant. Reason: The throttle valve varies only the cross-section of the line through which flow passes. The flow rate produced is dependent on the difference in the pressures upstream and downstream of the restriction. The flow rate through the throttle valve is thus dependent on pressure, in fact on both the supply and load pressures. The flow control valve incorporates a pressure compensator which maintains the internal pressure difference at a constant value. The flow rate is thus not dependent on the supply and load pressures.


C-49 Solution 12






C-50 Solution 12

Components list


Evaluation

Item no.

Qty.

Description

0Z1

1


0Z2, 1Z1, 1Z2

3

Pressure gauge

0V1

1


0V2

1

Shut-off valve

0V3

1

2-way flow control valve

1V

1


1A

1


7

Hose line

4

Branch tee

1

Stop-watch

Assemble and check the circuit. Close the shut-off valve 0V2 and the flow control valve 0V3. Now switch on the hydraulic power pack and set a system pressure of 50 bar by means of the pressure relief valve 0V1. Now open the shut-off valve 0V2 and also open the flow control valve until the piston rod advances. The measurements can now be carried out. p1Z1 = Pressure on piston side of cylinder p1Z2 = Pressure on annular piston side of cylinder p0Z2 = System pressure = 50 bar t→

Values table

= Cylinder advance-stroke time approx. 4 s Direction

p1Z1

p1Z2

t

Advance stroke

3.5 bar

5 bar

4.31 s

Return stroke

0 bar

4.5 bar

6.57 s


C-51 Solution 12

Cylinder dimensions: Piston area:

APN = 2.0 cm

2

Annular piston area:

APR = 1.2 cm

2

Cylinder stroke:

s = 0.2 m A PN 2 cm 2 = = 167 . ≈ 17 . A PR 12 . cm 2

Area ratio:

α=

Time ratio:

t adv 4.31 s = = 0.656 t ret 6.57 s

Force ratio:

F1 A ⋅p 2 cm 2 ⋅ 3.5 bar = PN 1Z1 = = 12 . <α F2 A PR ⋅ p 1Z 2 12 . cm ⋅ 5 bar

Flow rate during advance stroke: Piston side:

qPN = 9.28

Annular piston side:

s

qPN = A PN ⋅

t adv

= 2 cm 2 ⋅

20 cm 4.31 s

cm 3 cm 3 l = 557 ≈ 0.6 s min min

qPR = A PR ⋅

qPR = 5.57

s t adv

= 12 . cm 2 ⋅

20 cm 4.31 s

cm 3 cm 3 l = 334 ≈ 0.3 min min s

Flow rate during return stroke: Annular piston side:

qPR = A PR ⋅

qPR = 3.65


s t ret

= 12 . cm 2 ⋅

20 cm 6.57 s

cm cm 3 l = 219 = 0.2 = qFCV min min s

C-52 Solution 12

Conclusions

If the same pressure acts on a larger area (APN), this produces a larger force (F1). Mathematical proof: Given

p 1Z1 = p 1Z 2

and

p 1Z1 =

we optain

F1 A PR = =α F2 A PR

it follows:

F1 = α ⋅ F2

F1 F und p 1Z 2 = 2 A PN A PR

Since α > 1, F1 > F2, and the cylinder advances. The lower travel pressure p1Z1 acts on the larger area APN, producing a greater force F1. Only when the ratio of travel pressure to back pressure becomes equal to the area ratio a is an equilibrium of forces achieved, causing the piston to stop. Mathematical proof: F1 A ⋅p = PN 1Z1 F2 A PR ⋅ p 1Z 2

Für

F1 = F2

gilt

p 1Z 2 A = PN = α p 1Z1 A PR

As long as

p 1Z 2 < α , the piston will advance. p 1Z1


C-53 Solution 12

The available cylinder force is: F = F1 − F2 = (A PN ⋅ p 1Z1 ) − (A PR ⋅ p 1Z 2 ) F = (2 cm 2 ⋅ 3.5 bar) − (12 . cm 2 ⋅ 5 bar) F = (7 kp − 6 kp) = 1 kp = 10 N A comparison of a simple cylinder control circuit and a differential circuit reveals the following differences: System

1. Advance-stroke speed vadv 2. Return-stroke speed vret 3. Advance-stroke time tadv 4. Return-stroke time tret



Adjustable on FCV ≈ qFCV

Greater than set on FCV > qFCV

Greater than advance-stroke speed > vadv

Less than advance stroke speed < vadv (with α < 2)

Adjustable on FCV ≈ qFCV

Less than set on FCV < qFCV

Less than Advance-stroke time < tadv

Greater than Advance-stroke time > tadv (with α < 2)

With an area ratio of α = 2, the advance-stroke and return-stroke speeds are the same. Reason: The flow rate required on the piston side is double that necessary on the piston rod side. During the advance stroke, this flow rate is supplied from the pump and the annular piston side. During the return stroke, only supply from the pump is available. The return-stroke speed is produced by this. The advance-stroke and return-stroke speeds can be the same only when the area ratio α = 2.


General comparison

C-54 Solution 12

Below is the mathematical proof of this: Simple cylinder control circuit and differential circuit

APN

APR

APN

APR

vadv

qPN

qPR

qFCV

vadv

vPN

qPR

qFCV

Basic equations:

Area ratio:

α=

A PN A PR

Speed:

v=

q A

Travel time:

t=

s v


C-55 Solution 12

1.

Advance-stroke speed

Simple cylinder control circuit:

v adv =

q qPN = FCV A PN A PN

With differential circuit:

v adv =

qPN A PN

Flow rate on piston side:

qPN = qFCV + qPR

Given

qPN A = PN = α qPR A PR

we obtain:

qPR =

It follows:

qPN = qFCV +

1 ⋅ qPN α

qPN ⋅ (1 − qPN ⋅

1 ⋅ qPN α

1 ) = qFCV α

α −1 = qFCV α

qPN =

α ⋅ qFCV α −1

The advance-stroke speed with a differential circuit is thus: Vadv =

For α = 2

q α ⋅ FCV α − 1 A PN

Vadv = 2 ⋅

qFCV , A PN

and is thus twice as high as with the simple cylinder control circuit.


C-56 Solution 12

2.

Return-stroke speed

Simple cylinder control circuit: v ret =

q q qPR = FCV = α ⋅ FCV A PR A PR A PN

v ret = α ⋅ v adv Since α > 1,

v ret > v adv

With differential circuit: v ret =

v ret v adv

q q qPR = FCV = α ⋅ FCV A PR A PR A PN

qFCV A PN = α −1 = α qFCV ⋅ α − 1 A PN α⋅

v ret = (α − 1) ⋅ v adv For α = 2,

v ret = v adv


C-57 Solution 12

3.

Advance-stroke time s v

In general::

t=

Simple cylinder control circuit::

t adv =

With differential circuit::

t adv =

With α = 2,

t adv =

4.

s

=

qFCV A PN

A PN ⋅ s qFCV

s α qFCV ⋅ α − 1 A PN

=

α − 1 A PN ⋅ s ⋅ α qFCV

1 A PN ⋅ s ⋅ 2 qFCV

Return-stroke time

Simple cylinder control circuit:

t ret =

With differential circuit::

t ret =

t ret t adv

A ⋅s s = PN q α ⋅ qFCV α ⋅ FCV A PN A ⋅s s = PN q α ⋅ qFCV α ⋅ FCV A PN

A PN ⋅ s α ⋅ qFCV 1 = = α − 1 A PN ⋅ s α − 1 ⋅ α qFCV

t ret 1 = t adv α − 1

With α = 2,


t ret = t adv

C-58 Solution 12

The evaluation can thus be expressed as follows: Mathematical comparison

System



qFCV A PN

α qFCV ⋅ α − 1 A PN

2. Return-stroke speed vret

α ⋅ v adv

(α − 1) ⋅ v adv

3. Advance-stroke time tadv

A PN ⋅ s qFCV

α − 1 A PN ⋅ s ⋅ α qFCV

4. Return-stroke time tret

1 ⋅ t adv α

1 ⋅ t adv α −1

1. Advance-stroke speed vadv

Alternative solutions giving identical advance-stroke and return-stroke speeds: Differential circuit with 4/3-way valve with special mid-position

Equal-speed cylinder controlled by 4/2-way valve


C-59 Solution 13

Drilling machine



C-60 Solution 13


System pressure p = 50 bar (5 MPa) Pump safety valve pmax = 60 bar (6 MPa)


C-61 Solution 13

Item no.

Qty.

0Z1

1


0Z2, 1Z1, 1Z2, 1Z3

4

Pressure gauge

0V

1


1V1

1


1V2

1

Pressure regulator

1V3

1

Non-return valve

1V4

1

Shut-off valve

1A

1


1V5

1

One-way flowcontrol valve

15

Hose line

5

Branch tee


Description

Components list

C-62 Solution 13


In the first task in the exercise, the travel pressures are measured; the inlet pressure can be set to 15 bar (as shown on p1Z2) only after the piston has reached its forward end position or is opposed by a resistance. This is demonstrated by task 2 (piston in forward end position). This task also shows that the pressure regulator maintains a pressure of 15 bar even without through-flow. The valves 1V3 and 1V4 provide a bypass of the pressure regulator to allow a faster return stroke to be achieved. If the advance stroke is opposed by a resistance, as in task 3, a flow pressure of only 12 - 15 bar is achieved, despite the system pressure of 50 bar. By closing the throttle valve 1V5, it is possible to increase the counter pressure until the pressure gauge p1Z2 shows 15 bar; the piston will then stop, i.e. the pressure regulator will close. In task 5, it is demonstrated that increased counter pressure during the return stroke causes the valve to the tank to open, resulting in only the set pressure of 15 bar being attained. The piston can be pushed into the retracted end position. With the piston in this position, as in task 6, the 15 bar pressure is initially maintained. Due to internal leakage within the valve, the pressure then falls below 15 bar, causing the pressure regulator to switch from A - T to P - A. As no pump delivery is reaching the line to the pressure regulator via the 4/3-way valve, the pressure falls to 0 bar.

In practice, a pressure relief valve with bypass must be used in place of the one-way flow control valve 1V5. This prevents the high pressures which would arise upstream of the one-way flow control valve due to pressure intensification during the advance stroke of the piston. A oneway flow control valve has been used in this case to simplify the circuit configuration. Excessive pressures cannot arise in this case due to the fact that the system is being operated with reduced pressure.


C-63 Solution 13

Evaluation

Measured are: p1Z1 = Pressure upstream of pressure regulator p1Z2 = Pressure upstream of cylinder p1Z3 = Pressure downstream of cylinder Cases of examination: 1. Piston advance stroke 2. Piston advanced to end position with setting p1Z2 = 15 bar. 3. Piston advance stroke with counter pressure setting, p1Z3 = 20 bar. 4. Piston advanced to end position 5. Piston advance stroke with shut-off valve closed 6. Piston advanced to end position with shut-off valve closed Cases of examination

p1Z1

p1Z2

p1Z3

1. Advance stroke

5 bar

2 bar

1 bar

2. End position

50 bar

15 bar

0 bar

3. Advance stroke with counter pressure

49 bar

13 bar

20 bar

4. End position

50 bar

16 bar

0 bar

5. Advance stroke with pressure regulator

49 bar

14 bar

20 bar

6. End position

50 bar

16 bar

0 bar

p1Z1

p1Z2

p1Z3

1. Return stroke

4 bar

7 bar

16 bar

2. End position

0 bar

0 bar

50 bar

3. Return stroke with counter pressure

6 bar

10 bar

18 bar

4. End position

0 bar

0 bar

50 bar

5. Return stroke with pressure regulator

0 bar

26 bar

46 bar

6. End position

0 bar

0 bar

50 bar

Cases of examination


Advance stroke

Return stroke

C-64 Solution 13

Conclusions

Pressure regulators are used in cases when a secondary circuit with a constant but lower pressure is required in addition to a primary circuit. Note that increased pressures occur which act on port A of the pressure regulator. These pressures must be discharged to the tank.


C-65 Solution 14

Bulkhead door

Circuit diagram, hydraulic, without counter-holding




C-66 Solution 14

Practical assembly, hydraulic, with counter-holding

In inlet line p = 10 bar (1 MPa)




C-67 Solution 14

Item no.

Qty.

Description

0Z1

1


0Z2, 1Z1, 1Z2

3

Pressure gauge

0V1, 1V3

2


0V2

1

Shut-off valve

1V1

1

4/2-way valve

1V2

1


1A

1


1Z3

1

Loading weight

1V4

1

Non-return valve

12

Hose line

4

Branch tee

1

Stop-watch

Assemble and check the circuit. Mount the cylinder 1A on the profile in such a way that it can advance downwards. First close the shut-off valve 0V2. Switch on the hydraulic power pack and then use the pressure relief valve 0V1 to set a system pressure of 50 bar. Open the shut-off valve and adjust the pressure relief valve 1V2 in such a way that the piston rod advances in approx. 5 sec. The throttle valve setting should be retained while manipulating the circuit using the weight 1Z3, with counter-holding provided by the pressure relief valve 1V3. For the return stroke, a non-return valve 1V4 is required as a bypass for the pressure relief valve. After the measurements have been completed, first remove the weight and then retract the cylinder. Now depressurise the circuit by closing the shut-off valve and then opening the pressure relief valve 1V3. Dismantle the circuit only when the pressure has fallen to zero, as shown by the pressure gauge 1Z2.


Components list


C-68 Solution 14

Evaluation



p1Z1 = Cylinder travel pressure p1Z2 = Cylinder back pressure p0Z2 = System pressure Values table

Conclusions

p0Z2

p1Z1

p1Z2

t→


50 bar

0 bar

0 bar

5.0 s


50 bar

0 bar

0 bar

0.8 s


50 bar

2 bar

10 bar

4.6 s


50 bar

7 bar

10 bar

5.3 s


The travel time becomes shorter as the load increases. Reason: The piston is pulled out by the load. Without counter-holding, the movement is uncontrolled and jerky. A constant advance-stroke speed is obtained only with counter-holding. The generation of a counter pressure clamps the piston hydraulically. The travel and back pressures remain constant, which means that the travel speed also remains constant. A circuit with counter-holding is advisable both with and without a load. It is also possible to adjust the counter-holding to suit the load.


C-69 Solution 15

Ferry loading ramp

Circuit diagram, hydraulic, with counter-holding and flow control valve in inlet line

In Inlet line p = 10 bar (1 MPa)




C-70 Solution 15

Practical assembly, hydraulic, with flow control valve in outlet line




C-71 Solution 15

Item no.

Qty.

Description

0Z1

1


0Z2, 1Z1, 1Z2

3

Pressure gauge

0V1, 1V4

2


0V2

1

Shut-off valve

1V1

1

4/3-way valve

1V3

1

Flow control valve

1V2, 1V5

2

Non-return valve

1A

1


1Z3

1

Loading weight

13

Hose line

4

Branch tee

1

Stop-watch

Assemble and check the circuit. Mount the cylinder 1A on the profile plate in such a way that it can advance downwards. First close the shutoff valve 0V2. Switch on the hydraulic power pack and then use the pressure relief valve 0V1 to set a system pressure of 50 bar. Open the shut-off valve and adjust the pressure relief valve 1V3 in such a way that the piston rod advances in approx. 5 sec. The flow control valve setting should be retained throughout the series of measurements. Only the circuit should be modified. Dismantle the circuit only when the pressure has fallen to zero, as shown by the pressure gauge 1Z2.


Components list


C-72 Solution 15

Evaluation



p1Z1 = Cylinder travel pressure p1Z2 = Cylinder back pressure p0Z2 = System pressure Flow control valve in inlet line

Flow control valve in outlet line

Conclusions

p0Z2

p1Z1

p1Z2

t→

Without load without counter-holding

50 bar

0 bar

0 bar

5s


50 bar

0 bar

0 bar

0.6 s


50 bar

3 bar

10 bar

5s


50 bar

8 bar

10 bar

5s

p0Z2

p1Z1

p1Z2

t→

Without load

50 bar

48 bar

77 bar

5s

With load

50 bar

48 bar

84 bar

3.1 s


Load

Without counter-holding, the piston rod is pulled out by the load. It advances jerkily. With counter-holding, the same speed is achieved with and without a load. If, however, the flow control valve is installed in the outlet line to provide counter-holding, very high pressures will occur on the outlet side. This is often unacceptable in practice. A suitable circuit is thus one with a flow control valve in the inlet line and counter-holding by means of a pressure relief valve in the outlet line.


C-73 Solution 16

Skip handling


p = 30 bar (3 MPa)

p = 50 bar (5 MPa)




C-74 Solution 16

Components list


Item no.

Qty.

Description

0Z1

1


0Z2, 1Z1, 1Z2

3

Pressure gauge

0V, 1V3, 1V4

3


1V1

1


1V2, 1V5

2

Non-return valve

1A

1


10

Hose line

4

Branch tee

Assemble the control circuit in accordance with the circuit diagram. Ensure that the non-return valves are installed correctly. Open the pressure relief valves fully. If a sufficient number of pressure relief valves are not available, the system pressure can also be set on the pressure relief valve of the hydraulic power pack. Check the circuit and then switch on the hydraulic power pack. Now set the system pressure to 50 bar. When the directional control valve 1V1 is actuated, the cylinder advances and retracts at maximum speed. The travel motion can be slowed down by closing the two pressure relief valves 1V3 and 1V4. Adjustment is carried out using the pressure relief valve in the outlet line in each case. The non-return valves 1V2 and 1V5 are used to bypass the pressure relief valves fitted in each inlet line. The counter-holding pressures are shown on the pressure gauges 1Z1 and 1Z2. If the non-return valves are installed incorrectly, the travel speed will not change even when the pressure relief valves are closed. With the pressure relief valves fully closed, the cylinder will no longer retract due to the pressure intensification effect present.

Before the circuit is dismantled, the pressure relief valves must once again be fully opened to ensure that no pressure is trapped.


C-75 Solution 16

Hydraulic clamping on both sides is provided by two counter-holding circuits with pressure relief valves. A non-return valve is required as a bypass in each direction. Take account of the cylinder area ratio when setting the pressure.

Conclusions

In practice, skip handling is controlled using proportional valves. Dynamic adjustment of the proportional valves allows better control of fast and slow travel motions.

Remark


C-76 Solution 16


C-77 Solution 17

Bonding press

Practical assembly, hydraulic With 3-way pressure regulator

With pressure relief valve

p = 30 bar (3 MPa)

p = 30 bar (3 MPa)

p = 50 bar (5 MPa)

pmax = 60 bar (6 MPa)


p = 50 bar (5 MPa)


C-78 Solution 17

Components list

Item no.

Qty.

Description

0Z1

1


0Z2, 1Z1, 1Z2

3

Pressure gauge

0V, 1V4

2


1V1

1


1V2

1

Shut-off valve

1V2

1

Pressure regulator

1A

1


7

Hose line

5

Branch tee


In the case of the circuit with the pressure regulator, the shut-off valve must be opened to retract the piston rod. Due to the pressure intensification effect, the system pressure of 50 bar is not sufficient to open the pressure regulator from A to T.

Conclusions

If a pressure relief valve is fitted in the bypass, the overall system pressure will fall to 30 bar during the advance stroke. If a pressure regulator is used, the system pressure of 50 bar is maintained, and only the cylinder is supplied with the reduced pressure of 30 bar. This allows further actuators to be supplied with full system pressure by the same hydraulic power pack. Check, however, that the pump delivery is sufficient for this. The pressure relief valve gives an advantage in this application, since, in the case of long standstill periods with the directional control valve actuated, the pump need only develop the set pressure of 30 bar.


C-79 Solution 18

Assembly device


p = 20 bar (2 MPa)

p = 30 bar (3 MPa)

q = 1 l/min

p = 50 bar (5 MPa)

p = 60 bar (6 MPa)


C-80 Solution 18

Displacement-step diagram

Time

Components

Step Designation

Description

Signal

4/3-way-valve *

Cylinder


Motor


* 4/3-way-valve

Switching position a:

Switching position 0:

Switching position b:


C-81 Solution 18

Item no.

Qty.

Description

0Z1

1


0V1, 1V2, 1V4

3


0V2

1

Flow control valve

0V3

1

Shut-off valve

1V1

1


1A1

1


1A2

1

Hydromotor

1V3, 1V5

2

Non-return valve

1Z1, 1Z2

2

Pressure gauge

16

Hose line

7

Branch tee

1

Flow sensor

Before assembling the circuit, set the flow control valve to a flow rate of 1 l/min. When assembling the circuit, ensure that the non-return valves are installed correctly, since otherwise pressure may become trapped. If a sufficient number of pressure relief valves is not available, the system pressure can be set on the pressure relief valve of the hydraulic power pack. Once the circuit has been assembled and checked, switch on the hydraulic power pack. The shut-off valve should be closed at this time. The system pressure of 50 bar can now be set on the pressure relief valve 0V1. The two other pressure relief valves should be closed. When the 4/3-way valve is actuated, fluid will first flow to cylinder 1A1, and the piston of this will advance. Motor 1A2 will begin to rotate only when the pressure relief valve 1V4 is opened. The return stroke is initiated by reversing the 4/3-way valve. The motor will then stop. A pressure will build up at the pressure relief valve 1V2. Cylinder 1A1 will retract when the pressure relief valve 1V2 is opened.


Components list


C-82 Solution 18

Conclusions

The most important steps in commissioning are as follows: 1. Presetting of flow rate 2. Assembly of circuit 3. Closing the pressure relief valves 4. Checking the circuit 5. Switching on the hydraulic power pack 6. Adjusting the pressure relief valves during the operation of the control circuit


C-83 Solution 19


Evaluation Schematic diagram

Piston force:

F1 = A PN ⋅ p 1 =

π 2 ⋅ D ⋅ p1 4

F1 =

π ⋅ 50 2 mm 2 ⋅ 50 bar 4

F1 =

π kp ⋅ 50 2 mm 2 ⋅ 50 4 cm 2

F1 =

π 50 2 mm 2 ⋅ 50 kp ⋅ 4 100 mm 2

. kp = 9817.5 N F1 = 98175 F1 = 9.8 kN


C-84 Solution 19

Counter force:

F2 = A PR ⋅ p 2 =

π ⋅ (D 2 − d 2 ) ⋅ p 2 4

F2 =

π ⋅ (50 2 − 25 2 ) mm 2 ⋅ 6 bar 4

F2 =

π kp ⋅ (50 2 − 25 2 ) mm 2 ⋅ 6 4 cm 2

F2 =

π 1875 mm 2 ⋅ 6 kp ⋅ 4 100 mm 2

F2 = 88.36 kp = 883.6 N F2 = 0.9 kN

Press-fitting force: F = F1 − F2 = 9.8 kN − 0.9 kN F = 8.9 kN

Press-fitting time:

π 2 ⋅D ⋅ s V A ⋅s 4 = t = = PN q q q t=

t=

t=

π 50 2 mm 2 ⋅ 250 mm ⋅ l 4 5 min π 5 2 cm 2 ⋅ 25 cm ⋅ 4 5000 cm 3 60 s π 625 cm 3 ⋅ 60 s π 625 ⋅ 60 s ⋅ = ⋅ 4 4 5000 5000 cm 3

t = 5.89 s ≈ 6 s


C-85 Solution 20

Tipping container


p = 50 bar (5 MPa)



C-86 Solution 20

Circuit diagram, electrical

S1 = “Up” pushbutton S2 = “Down” pushbutton

Components list, hydraulic

Components list, elektrisch

Item no.

Qty.

0Z

1


0V

1


1V1

1


1V2

1

Non-return valve, hydraulically piloted

1A

1


6

Hose line

2

Branch tee

Qty.

Description

Item no.

Description

1

Signal input unit

1

Relay, 3-fold

1

Cable set

1

Power supply unit


C-87 Solution 20

A piloted non-return valve is used to protect the tipping container against undesired lowering. A 4/3-way valve with a mid-position in which A, B and T are connected and P is closed is used in order to ensure that the non-return valve closes when the electrical control circuit is switched off. This 4/3-way valve relieves ports A and B in its mid-position.


Once the electrical and hydraulic circuits have been assembled and checked, actuate push-button S1. This causes the cylinder piston rod to travel to its forward end position (filling position). When the push-button S1 is released, a spring force causes the 4/3-way valve to switch to its mid-position. The load acting on the piston rod now causes the pilotoperated non-return valve to close, which prevents the piston rod from being pushed back. When the push-button S2 is actuated, the 4/3-way valve reverses. The pressure which builds up in the line from port B of the valve causes the non-return valve to open and the piston rod of the cylinder travels into its retracted end position (emptying position). The two push-buttons S1 and S2 each actuate one normally-open and one normally-closed contact, which are connected together in such a way that no movement occurs if the push-buttons are actuated simultaneously.

It is also possible to carry out these exercises using the 4/3-way valve with recirculating mid-position which is included in the equipment set. Due to the inherent characteristics of this valve, internal leakage losses can occur which will cause the piloted non-return valve to close. The electrical circuit diagram incorporates an interlock between current paths 1 and 2. This ensures that the control circuit will work correctly even in the case of operator error.


Conclusions

C-88 Solution 20


Festo Hydraulic Basic Level Manual

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